On the sum of power laws
Many real-life data sets have power laws or Zipfian distributions. An integer-valued random variable X follows a power law with parameter a if P(X=k) is proportional to k-a. Panos asked what the sum of two power laws was. He cites Wilke at al. who claim that the sum of two power laws X and Y with parameters a and b is a power law with parameter min(a,b).
I relate this problem to the sum of exponentials. Any engineer knows that if a>b, then eat + ebt will be approximately eat for t sufficiently large.
Hence, I think that the sum of power law distributions X and Y is a power law distribution with parameter min(a,b) if you are only interested in large values of k in P(X+Y=k).
For extra credit, help me solve this problem. Suppose that I have two power laws with the same parameter. Is their sum a power law with the same parameter? (I predict it does not!)
Egghe showed in The distribution of N-grams that even if the words follow a power law, the n-grams won’t!
Disclaimer: Yes, I am being lazy. I could work it out.
Montreal, Canada 
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(1) Mandelbrot has proposed a generalization of Zipf’s Law. (2) Randomly generated strings follow Zipf’s Law, so some people argue that in some cases it is a statistical artifact.
http://en.wikipedia.org/wiki/Zipf%27s_law
Comment by Peter Turney — 25/1/2008 @ 13:17
It seems that power law is really the same as Pareto distribution . This paper gives some closed formulas for distributions of sums of Pareto, which are themselves not Pareto
If two power laws have different parameters, as you go to infinity, odds of encountering the one with higher a becomes vs. one with lower a goes to 0, so I also expect that for large values, heavier tail distribution will dominate
BTW, I also wondered about distribution of bigrams when unigrams are power-law distributed, David Cantrell in sci.math gave an approximate formula for the cdf involving Lambert’s W function
http://groups.google.com/group/sci.math/browse_thread/thread/8de7cee65f65ff70/810470b85f36523b?lnk=st&q=group%3Asci.math#810470b85f36523b
Comment by Yaroslav Bulatov — 28/1/2008 @ 21:25
Yaroslav,
Thanks, very useful!
- Panos
Comment by Panos Ipeirotis — 31/1/2008 @ 21:34